$\newcommand\CC{\mathbb C}\newcommand\ZZ{\mathbb Z}\newcommand\ad{\mathsf{ad}}\newcommand\Ext{\operatorname{Ext}}$ Suppose that $G$ is a finite group and that it acts on a finite dimensional complex vector space $V$.
For each $g\in G$ we may consider the subspace $V_g$ of $V$ spanned by the eigenvectors of $g$ corresponding to eigenvalues different from $1$, and put $d(g)=\dim V_g$. We pick any nonzero linear map $\omega_g:\Lambda^{d(g)}V_g\to\CC$.
If $h\in G$ then the map $x\in \Lambda^{d(g)}V_{hgh^{-1}}\mapsto \omega_g(h^{-1}v)\in\CC$ makes sense, and we write it $h\cdot\omega_g$. Since it is not zero, there is a nonzero scalar $\lambda(h,g)\in\CC^\times$ such that $h\cdot\omega_g=\lambda(h,g)\omega_{hgh^{-1}}$. Associativity of the group action implies that $$\lambda(gh,k)=\lambda(h,k)\lambda(g,hkh^{-1})$$ for all $g$, $h$, $k\in G$. This means that the function $\lambda:G\times G\to\CC^\times$ is a $1$-cocycle in the complex which computes $\Ext_{\ZZ G}^\bullet(\ZZ G^\ad,\CC^\times)$, when we use a bar resolution of the $G$-module $\ZZ G^\ad$ (which is the permutation $G$-module constructed from the conjugation action of $G$ on itself) We may therefore take its class $[\lambda]$ in $\Ext^1_{\ZZ G}(\ZZ G^\ad,\CC^\times)$.
This class depends only on the representation and not on the choice of the $\omega_g$'s, so we write it $c(V)$.
The long exact sequence corresponding to the exponential sequence $0\to\ZZ\to\CC\to\CC^\times\to0$ allows us to identify that $\Ext^1$ with $\Ext^2_{\ZZ G}(\ZZ G^\ad,\ZZ)$. If $CG$ is the set of conjugacy classes and for each $c\in CG$ we let $\ZZ c$ be the permutation module corresponding to the conjugation action on $c$, we have $\ZZ G^\ad=\bigoplus_{c\in CG}\ZZ c$. If $G_c$ denotes the centralizer of an element of $c$, some form of Shapiro's lemma allows us to identify $\Ext^2_{\ZZ G}(\ZZ c,\ZZ)$ with $\Ext^2_{\ZZ G_c}(\ZZ,\ZZ)$ and then our class $[\lambda]$ is an element of $$\bigoplus_{c\in CG}H^2(G_c,\ZZ).$$ This can be written a bit more canonically as $$\left(\bigoplus_{g\in G}H^2(C_g,\ZZ)\right)^G$$ with now $C_g$ the centralizer of $g\in G$ and $G$ acting on the direct sum in the only sensible way. If $G$ is abelian, this can be identified further to $\ZZ G\otimes H^2(G,\ZZ)$.
Question: What is this cohomology class $c(V)$?
This reminds me of Burghelea's description of the cyclic homology of the group algebra, so it might be some form of Chern class (if some vector bundle constructed from V on the free loop space of BG?).
This class shows up when one computes the Hochschild cohomology of the cross product algebra $S(V)\rtimes G$, but explaining how would make this post too much longer. It appears as an obstruction to making this nice :-/
On triviality: suppose there is a central element $z$ in $G$ which acts by multiplication by a scalar different from $1$. In that case $c=\{z\}$ is a conjugacy class, $G_c=G$, $\lambda(g,z)=\det g$ for all $g$, and the component of $c(V)$ in $\Ext^1_{\ZZ G}(\ZZ c,\CC^\times)=\hom(G,\CC^\times)$ is the determinant map. If the representation does not land in $SL(V)$, then this is not zero. Examples of representations like this are the geometric representation of Weyl groups which have nontrivial center, in which case the longest element is central and acts as $-1$ —see this question and its answers for the list of which types work— or abelian groups acting with some element not having $1$ in its spectrum.
In the other direction, if the representation respects a symplectic form $\omega$ on $V$, then all the $d(g)$ are even and we may take $\omega_g$ to be the restriction of the $\tfrac12d(g)$th power $\omega^{d(g)/2}$ to $V_g$. In this case, $\lambda(h,g)=1$ identically, so $c(V)=0$.
Note. If $e$ is the exponent of $G$ and $\Omega$ is any set of $e$th roots of unity, we can consider the subspace $V_g^\Omega$ spanned by eigenvectors of $g$ corresponding to eigenvalues in $\Omega$, and proceeding as above we get a class $c_\Omega(V)$ in the same direct sum. The class above corresponds to taking $\Omega$ the set of all $e$th roots of unity different from $1$. The assigment $V\mapsto\bigoplus_{g\in G}V_g^\Omega$, for $c$ a conjugacy class of $G$, is a nice functor into Yetter-Drinfeld modules, by the way.
